Question

Which of the following option(s) is/are correct for the ground state of a hydrogen atom?

• A. Linear Stark effect is zero
• B. It has definite parity
• C. Spin-orbit coupling is zero
• D. Hyperfine splitting is zero

Hint:

For the ground state of hydrogen, there is no linear Stark effect due to spherical symmetry, it has definite parity, and spin-orbit coupling is zero due to the absence of orbital angular momentum.

Answer 1

The Correct Option is A, B, C

Let's analyze each option:

1. Linear Stark effect is zero:
In the ground state of a hydrogen atom (which is the \( 1s \) state), the electric dipole moment is zero because the electron is spherically symmetric. Hence, the linear Stark effect, which is proportional to the electric dipole moment, is zero in this case.
Therefore, (A) is correct.

2. It has definite parity:
The parity of a quantum state is determined by the wavefunction. The wavefunction for the hydrogen atom in the ground state (\( 1s \) state) is spherically symmetric, and therefore, the parity of the ground state is well-defined.
Therefore, (B) is correct.

3. Spin-orbit coupling is zero:
The ground state of the hydrogen atom is in the \( 1s \) orbital, which has no orbital angular momentum (\( l = 0 \)). Since spin-orbit coupling is proportional to the interaction between the electron’s spin and its orbital angular momentum, and \( l = 0 \) in the ground state, the spin-orbit coupling is zero.
Therefore, (C) is correct.

4. Hyperfine splitting is zero:
Hyperfine splitting arises due to the interaction between the nuclear spin and the electronic spin. In the case of the hydrogen atom, this interaction is not exactly zero, but it is extremely small. However, it is not exactly zero, so this statement is incorrect.
Therefore, (D) is incorrect.